Question

Let $x = \sum_{i=1}^{10} \frac{1}{10\sqrt{3}}[\frac{1}{1+(\frac{i}{10\sqrt{3}})^2}]$ and $y = \sum_{i=0}^{9} \frac{1}{10\sqrt{3}}[\frac{1}{1+(\frac{i}{10\sqrt{3}})^2}]$, then-

• A. $x < \frac{\pi}{6}$
• B. $y > \frac{\pi}{6}$
• C. $x+y < \frac{\pi}{3}$
• D. $x+y > \frac{\pi}{3}$

Answer 1

Answer: (A), (B), (D)

x < \frac{\pi}{6}, y > \frac{\pi}{6}, x+y > \frac{\pi}{3}

Answer 2

Solution Explanation:

Write

$$\Delta x=\frac{1}{10\sqrt{3}}$$

and notice that

$$x=\sum_{i=1}^{10}\Delta x\, \frac{1}{1+\left(\frac{i}{10\sqrt{3}}\right)^2},\quad y=\sum_{i=0}^{9}\Delta x\, \frac{1}{1+\left(\frac{i}{10\sqrt{3}}\right)^2}.$$

These are the right‐ and left–Riemann sums, respectively, for

$$I=\int_{0}^{\frac{1}{\sqrt{3}}}\frac{1}{1+x^2}\,dx.$$

Since

$$\int_{0}^{\frac{1}{\sqrt{3}}}\frac{dx}{1+x^2}=\arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6},$$

and because the function $f(x)=\frac{1}{1+x^2}$ is decreasing on $[0,\frac{1}{\sqrt{3}}]$, the right sum underestimates and the left sum overestimates the integral. Hence:

• $x<\frac{\pi}{6}$
• $y>\frac{\pi}{6}$.

Also, the trapezoidal rule is given by $\frac{x+y}{2}$. For a concave function (here, $f$ is concave on $[0,\frac{1}{\sqrt{3}}]$), the trapezoidal rule gives an overestimate. Therefore:

$$\frac{x+y}{2}>\frac{\pi}{6}\quad\Longrightarrow\quad x+y>\frac{\pi}{3}.$$